“JUST THE MATHS”
UNIT NUMBER
10.3
DIFFERENTIATION 3
(Elementary techniques of differentiation)
by
A.J.Hobson
10.3.1
10.3.2
10.3.3
10.3.4
Standard derivatives
Rules of differentiation
Exercises
Answers to exercises
UNIT 10.3 - DIFFERENTIATION 3
ELEMENTARY TECHNIQUES OF DIFFERENTIATION
10.3.1 STANDARD DERIVATIVES
In Unit 10.2, reference was made to the use of a table of standard derivatives and such a
table can be found in the appendix at the end of this Unit.
However, for the time being, a very short list of standard derivatives is all that is necessary
since other derivatives may be developed from them using techniques to be discussed later
in this and subsequent Units.
f (x)
a const.
xn
sin x
cos x
ln x
f 0 (x)
0
nxn−1
cos x
− sin x
1
x
Note:
In the work which now follows, standard derivatives may be used which have not, here, been
obtained from first principles; but the student is expected to be able to quote results from a
table of derivatives including those for which no proof has been given.
10.3.2 RULES OF DIFFERENTIATION
(a) Linearity
Suppose f (x) and g(x) are two functions of x while A and B are constants. Then
d
d
d
[Af (x) + Bg(x)] = A [f (x)] + B [g(x)].
dx
dx
dx
Proof:
The left-hand side is equivalent to
[Af (x + δx) + Bg(x + δx)] − [Af (x) + Bg(x)]
δx→0
δx
lim
"
#
"
f (x + δx) − f (x)
g(x + δx) − g(x)
= A lim
+ B lim
δx→0
δx→0
δx
δx
1
#
=A
d
d
[f (x)] + B [g(x)].
dx
dx
The result, so far, deals with a “linear combination” of two functions of x but is easily
extended to linear combinations of three or more functions of x.
EXAMPLES
1. Write down the expression for
dy
dx
in the case when
y = 6x2 + 2x6 + 13x − 7.
Solution
Using the linearity property, the standard derivative of xn , and the derivative of a
constant, we obtain
d
d
d
d
dy
= 6 [x2 ] + 2 [x6 ] + 13 [x1 ] −
[7]
dx
dx
dx
dx
dx
= 12x + 12x5 + 13.
2. Write down the derivative with respect to x of the function
5
− 4 sin x + 2 ln x.
x2
Solution
d 5
− 4 sin x + 2 ln x
dx x2
=
i
d h −2
5x − 4 sin x + 2 ln x
dx
2
= −10x−3 − 4 cos x +
x
−10
2
= 3 − 4 cos x + .
x
x
2
(b) Composite Functions (or Functions of a Function)
(i) Functions of a Linear Function
Expressions such as (5x + 2)16 , sin(2x + 3) and ln(7 − 4x) may be called “functions of a
linear function” and have the general form
f (ax + b),
where a and b are constants. The function f (x) would, of course, be the one obtained on
replacing ax + b by a single x; hence, in the above illustrations, f (x) would be x16 , sin x and
ln x, respectively.
Functions of a linear function may be differentiated as easily as f (x) itself on the strength
of the following argument:
Suppose we write
y = f (u) where u = ax + b.
Suppose, also, that a small increase of δx in x gives rise to increases (positive or negative)
of δy in y and δu in u. Then:
dy
δy
δy δu
= lim
= lim
.
dx δx→0 δx δx→0 δu δx
Assuming that δy and δu tend to zero as δx tends to zero, we can say that
dy
δy
δu
= lim
× lim
.
dx δu→0 δu δx→0 δx
That is,
dy
dy du
=
. .
dx
du dx
This rule is called the “Function of a Function Rule” or “Composite Function Rule”
or “Chain Rule” and has applications to a much wider class of composite functions than
has so far been discussed. But, for the moment we restrict the discussion to functions of a
linear function.
3
EXAMPLES
1. Determine
Solution
dy
dx
when y = (5x + 2)16 .
First, we write y = u16 where u = 5x + 2.
dy
= 16u15 and du
= 5.
du
dx
dy
Hence, dx
= 16u15 .5 = 80(5x +
Then,
2. Determine
dy
dx
2)15 .
when y = sin(2x + 3).
Solution
First, we write y = sin u where u = 2x + 3.
dy
= cos u and du
= 2.
du
dx
dy
Hence, dx = cos u.2 = 2 cos(2x
Then,
3. Determine
dy
dx
+ 3).
when y = ln(7 − 4x).
Solution
First, we write y = ln u where u = 7 − 4x.
dy
= u1 and du
= −4.
du
dx
dy
−4
= u1 .(−4) = 7−4x
.
Hence, dx
Then,
Note:
It is hoped that the student will quickly appreciate how the fastest way to obtain the derivative of a function of a linear function is to treat the expression ax + b initially as if it were
a single x; then, multiply the final result by the constant value, a.
(ii) Functions of a Function in general
The formula
dy
dy du
=
.
dx
du dx
is in no way dependent on the fact that the examples so far used to illustrate it have involved
functions of a linear function. Exactly the same formula may be used for the composite
function
4
f [g(x)],
whatever the functions f (x) and g(x) happen to be. All we need to do is to write
y = f (u) where u = g(x),
then apply the formula.
EXAMPLES
1. Determine an expression for
dy
dx
in the case when
y = (x2 + 7x − 3)4 .
Solution
Letting y = u4 where u = x2 + 7x − 3, we have
dy
dy du
=
.
= 4u3 .(2x + 7)
dx
du dx
= 4(x2 + 7x − 3)3 (2x + 7).
2. Determine an expression for
dy
dx
in the case when
y = ln(x2 − 3x + 1).
Solution
Letting y = ln u where u = x2 − 3x + 1, we have
dy
dy du
1
2x − 3
=
.
= .(2x − 3) = 2
.
dx
du dx
u
x − 3x + 1
3. Determine the value of
dy
dx
at x = 1 in the case when
y = 2 sin(5x2 − 1) + 19x.
Solution
Consider, first, the function 2 sin(5x2 − 1) which we shall call z.
dz
Its derivative is dx
, where z = 2 sin(5x2 − 1).
5
Let z = 2 sin u where u = 5x2 − 1; then,
dz
dz du
=
.
= 2 cos u.10x = 20x cos(5x2 − 1).
dx
du dx
Hence, the complete derivative is given by
dy
= 20x cos(5x2 − 1) + 19.
dx
Finally, when x = 1, this derivative has the value 20 cos 4 + 19, which is approximately
equal to 5.927, remembering that the calculator must be in radian mode.
Note:
Again, it is hoped that the student will appreciate how there is a faster way of differentiating
composite functions in general. We simply treat g(x) initially as if it were a single x, then
multiply by g 0 (x) afterwards.
For example,
i
d h 3 i
d h
sin x =
(sin x)3 = 3(sin x)2 . cos x = 3sin2 x. cos x.
dx
dx
10.3.3 EXERCISES
1. Determine an expression for
dy
dx
in the following cases:
(a)
y = 3x3 − 8x2 + 11x + 9;
(b)
y = 10 cos x + 5 sin x − 14x7 ;
(c)
y = (2x − 7)5 ;
(d)
5
y = (2 − 5x)− 2 ;
6
(e)
π
y = sin
− x ; that is, cos x;
2
(f)
y = cos(4x + 1);
(g)
y = ln(4 − 2x);
(h)
3x − 8
y = ln
.
6x + 2
2. Determine an expression for
dy
dx
in the cases when
(a)
y = (4 − 7x3 )8 ;
(b)
3
y = (x2 + 1) 2 ;
(c)
y = cos5 x;
(d)
y = ln(ln x).
3. If y = sin(cos x), evaluate
dy
dx
4. If y = cos(7x5 − 3), evaluate
at x = π2 .
dy
dx
at x = 1.
10.3.4 ANSWERS TO EXERCISES
1. (a)
9x2 − 16x + 11;
(b)
−10 sin x + 5 cos x − 98x6 ;
7
(c)
10(2x − 7)4 ;
(d)
7
25
(2 − 5x)− 2 ;
2
(e)
π
− x ; that is, − sin x;
− cos
2
(f)
−4 sin(4x + 1);
(g)
−2
2
or
;
4 − 2x
2x − 4
(h)
3
6
54
−
=
.
3x − 8 6x + 2
(3x − 8)(6x + 2)
2. (a)
−168x2 (4 − 7x3 )7 ;
(b)
1
3x(x2 + 1) 2 ;
(c)
−5cos4 x. sin x;
(d)
1
.
x ln x
3.
−1
4.
−35 sin 4 ∼
= 26.488
8
APPENDIX - A Table of Standard Derivatives
f (x)
f 0 (x)
a (const.)
0
xn
nxn−1
sin ax
a cos ax
cos ax
−a sin ax
tan ax
asec2 ax
cot ax
−acosec2 ax
sec ax
asec ax. tan ax
cosec ax
−acosec ax. cot ax
ln x
1/x
eax
aeax
ax
ax . ln a
sinh ax
a cosh ax
cosh ax
a sinh ax
tanh ax
asech2 ax
sech ax
−asech ax. tanh ax
cosech ax
−acosech ax. coth x
ln(sin x)
cot x
ln(cos x)
− tan x
ln(sinh x)
coth x
ln(cosh x)
tanh x
√
sin−1 (x/a)
1/ √(a2 − x2 )
cos−1 (x/a)
−1/ (a2 − x2 )
tan−1 (x/a)
a/(a2 + x2 )
√
−1
sinh (x/a)
1/ (x2 + a2 )
√
cosh−1 (x/a)
1/ (x2 − a2 )
tanh−1 (x/a)
a/(a2 − x2 )
9